What is the integral of zsqrt(z-9) ?

The integral of zsqrt(z-9) is 2/5 (z-9)^{5/2}+6(z-9)^{3/2}+C

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integral of zsqrt(z-9)

\int z z - 9 d z

\frac{2}{5} (z - 9)^{\frac{5}{2}} + 6 (z - 9)^{\frac{3}{2}} + C

Solution steps

\int z z - 9 d z

Apply u-substitution

: \int (u + 9) u d u

= \int (u + 9) u d u

Expand

(u + 9) u : u^{\frac{3}{2}} + 9 u

= \int u^{\frac{3}{2}} + 9 u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{\frac{3}{2}} d u + \int 9 u d u

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int 9 u d u = 6 u^{\frac{3}{2}}

= \frac{2}{5} u^{\frac{5}{2}} + 6 u^{\frac{3}{2}}

Substitute back

u = z - 9

= \frac{2}{5} (z - 9)^{\frac{5}{2}} + 6 (z - 9)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (z - 9)^{\frac{5}{2}} + 6 (z - 9)^{\frac{3}{2}} + C

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What is the integral of zsqrt(z-9) ?
The integral of zsqrt(z-9) is 2/5 (z-9)^{5/2}+6(z-9)^{3/2}+C